I have been thinking about doing a fun chicken-related experiment with solar but am having trouble believing my calculations are right! The idea is to tap solar energy to heat a bucket of water in the coop which would act as a heat dump to provide passive winter-time heat to a distant coop.
If I ran two hobby-sized encapsulated photovoltaic panels in series (each 6 volt, 50mA), like these it would generate (assume ideal conditions) 12V of DC current at 50 mA.
Then feed that current directly as a diversion load into a low-voltage water heating element (resistance=0.96 ohm) kind of like this .
So, according to Joules Law (Q=I^2*R*t) this should generate 0.0024 watts=(0.05amps^2*0.96ohms*1 second) per second under ideal conditions. This is ~8.64 watts/hr.
8.64 watts/hr is approximately equal to 29.5 BTU/hr.
Assume 8 hours of sunlight per day and that would be ~236 BTU/day
A BTU is the amount of energy required to raise 1 pound of water 1 degree Fahrenheit.
So, this setup would produce enough passive heat energy to raise a 5 gallon pail of water (which weighs 42lbs) ~5.6 degrees. This heat would be passively lost to the interior of the coop as the water cooled at night. The idea isn't to make the coop toasty, but rather to provide the marginal edge needed to keep chickens a bit more comfortable than they would be in a completely unheated coop.
But I am thinking my calculations must be off somewhere! Any electrical engineers in the house? Tell me where I've gone wrong.
If I ran two hobby-sized encapsulated photovoltaic panels in series (each 6 volt, 50mA), like these it would generate (assume ideal conditions) 12V of DC current at 50 mA.
Then feed that current directly as a diversion load into a low-voltage water heating element (resistance=0.96 ohm) kind of like this .
So, according to Joules Law (Q=I^2*R*t) this should generate 0.0024 watts=(0.05amps^2*0.96ohms*1 second) per second under ideal conditions. This is ~8.64 watts/hr.
8.64 watts/hr is approximately equal to 29.5 BTU/hr.
Assume 8 hours of sunlight per day and that would be ~236 BTU/day
A BTU is the amount of energy required to raise 1 pound of water 1 degree Fahrenheit.
So, this setup would produce enough passive heat energy to raise a 5 gallon pail of water (which weighs 42lbs) ~5.6 degrees. This heat would be passively lost to the interior of the coop as the water cooled at night. The idea isn't to make the coop toasty, but rather to provide the marginal edge needed to keep chickens a bit more comfortable than they would be in a completely unheated coop.
But I am thinking my calculations must be off somewhere! Any electrical engineers in the house? Tell me where I've gone wrong.